# Limiting reactant

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#### AwayFromReality

##### Full Member
7+ Year Member
So I have good grasp on the simple limiting reactant concept for the most part but I came across a problem which I wasn't sure of how to solve

A student starts with 138mg of 1,4 dimethoxybenzene (138 grams/mole) and treats is with 1.32 mL of anhydrous HNO3 (1.50 g/mL , 63 g/mole) dissolved into 2.61 mL anhydrous H2SO4. The final product is 91.5mg 2,5 dimethoxynitrobenzene (183g/mole). What is the percent yield?

I know we take actual yield over theoretical yield. I know actual yield is found by converting 91.5mg to moles. I know limiting reactant will give me the theoretical yield. What I don't know is how to figure out which one of the reactants is the limiting reactants just from the information above. Normally we have an equation which can tell us the relevant ratios and we can figure out the limiting reactant but we don't have that. So how should we approach this?

This is from TBR OChem Section VIII Practice Passage VII #43. The answer states that 1,4 dimethoxybenzene is the limiting reagent. I just don't know why.

EDIT: Maybe we are assuming a 1:1 product to reactant ratio for each reactant in which case 1,4 dimethoxybenzene will produce the lowest amount of product therefore is the limiting reactant. Does that sound right?

Thanks, that helps.

The limiting reagent is the one where, if it all went to product, you would get the smallest amount of product.
So, you have to convert mass amounts of reagents to moles, multiply by the relative stoichiometry and then compare.
[(g reactant) * (mol/g) * (mol product/mol reactant) = mol product (i.e. theoretical yield)) Do this for each reactant, and the one with the smallest amt of product is your limiting reagent.

In a rapid fire situation like the MCAT though, going with Hadi7183's suggestion of looking for the one with the smallest starting amount is a good idea.
Also, protip, when they've made one reagent really easy to calculate (138 mg & 138 g/mol) and the other not so easy to calculate in your head (1.32 mL * 1.50 g/mL * 1 mol/63 g), then that should be a big flashing arrow pointing to the one you're supposed to be working with. Ya know, since there isn't a calculator in the MCAT and concepts are more important...

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The limiting reagent is the one where, if it all went to product, you would get the smallest amount of product.
So, you have to convert mass amounts of reagents to moles, multiply by the relative stoichiometry and then compare.
[(g reactant) * (mol/g) * (mol product/mol reactant) = mol product (i.e. theoretical yield)) Do this for each reactant, and the one with the smallest amt of product is your limiting reagent.

In a rapid fire situation like the MCAT though, going with Hadi7183's suggestion of looking for the one with the smallest starting amount is a good idea.
Also, protip, when they've made one reagent really easy to calculate (138 mg & 138 g/mol) and the other not so easy to calculate in your head (1.32 mL * 1.50 g/mL * 1 mol/63 g), then that should be a big flashing arrow pointing to the one you're supposed to be working with. Ya know, since there isn't a calculator in the MCAT and concepts are more important...
Thanks. Yeah I did understand how to calculate but wasn't sure about the relative stoichiometry which is why I was confused. Now that I know about the nitration of benzene reaction I now know the 1:1 ratio which makes the question pretty simple.

I actually did think about how one reagent was far easier to calculate than the other one but seeing as how it was a TBR question and how they are known to make the questions a lot harder than the actual mcat, I didn't want to make that assumption. But you're right on the real mcat an assumption like that could definitely help.

Most of the questions I've seen either give you the reaction written out (so the stoichiometry is given) or have really straightforward reactions like above (so the stoich is obvious). When in doubt though, writing out a quick reaction won't hurt you.

I never did much with TBR, but all the prep materials I've seen do a pretty good job of keeping to the appropriate level of math/calculation. The harder-easier spectrum tends to run more along the lines of question wording, from what I've seen.

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"limiting reagent"

whole reaction
1A + 2B + 3c + 4D -----> 1E

broken down to individual cycles
1A ---->1E
2B ---->1E
3c ---->1E
4D ---->1E

put simply, if you had same amount of reactants, which reactant will run the reaction the least number of cycles.

if you had 4 moles of A,B,C,D, then A will run 4 cycles of reaction; B will run 2 cycles of reaction; C will run less than 2 cycles of reaction (3/2); D will run 1 cycle of reaction.

So, D is the limiting reagent

and limiting reagent dictate how many cycles of reaction is allowed to run to completion.
so, 4 moles of each means, D is limiting reagent, runs one cycle of reaction, producing only one mole of E.

"limiting reagent"

whole reaction
1A + 2B + 3c + 4D -----> 1E

broken down to individual cycles
1A ---->1E
2B ---->1E
3c ---->1E
4D ---->1E

put simply, if you had same amount of reactants, which reactant will run the reaction the least number of cycles.

if you had 4 moles of A,B,C,D, then A will run 4 cycles of reaction; B will run 2 cycles of reaction; C will run less than 2 cycles of reaction (3/2); D will run 1 cycle of reaction.

So, D is the limiting reagent

and limiting reagent dictate how many cycles of reaction is allowed to run to completion.
so, 4 moles of each means, D is limiting reagent, runs one cycle of reaction, producing only one mole of E.
Really appreciate your detailed explanation man. I knew how to solve limiting reagent problems, I just didn't know what the stoichoimetric ratio was in the nitrobenzene reaction above which is where I was stuck. But with other people's help I figured out the 1:1 ratio which made it obvious what the limiting reactant is. Thanks for your help.